Question: Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2}{x + 7} = \dfrac{-3x + 28}{x + 7}$
Multiply both sides by $x + 7$ $ \dfrac{x^2}{x + 7} (x + 7) = \dfrac{-3x + 28}{x + 7} (x + 7)$ $ x^2 = -3x + 28$ Subtract $-3x + 28$ from both sides: $ x^2 - (-3x + 28) = -3x + 28 - (-3x + 28)$ $ x^2 + 3x - 28 = 0$ Factor the expression: $ (x + 7)(x - 4) = 0$ Therefore $x = -7$ or $x = 4$ However, the original expression is undefined when $x = -7$. Therefore, the only solution is $x = 4$.